\section[$TM_z$ Radial Modes]{Transverse Magnetic $TM_z$ Radial Modes}
In order to obtain $TM_z$ modes, the following conditions of the vector potentials are
\begin{align}
\vec{A} &= A_z\hat{z}\\
\vec{F} &= 0
\end{align}
Using separation of variables, the magnetic vector potential is defined as
\begin{align}
A_z(\rho,\phi,z) &= {C_a}\left[\frac{R_A(\rho)}{\sqrt{Z_a}}\right]\Phi_A(\phi)Z_A(z)\label{eqn:Azcyl}\\
 &= {C_a}[I(\rho)]\Phi_A(\phi)Z_A(z)\label{eqn:Azcyl2}
\end{align}
where a $TM_\rho$ impedance is defined as,
\begin{align}
Z_a = \frac{k_{a\rho}}{j\omega\varepsilon'}\label{eqn:ZTMrho}
\end{align}
and where $C_a$ is a normalization constant.  The solutions must satisfy the wave equation
\begin{align}
(\nabla^2+k^2)A_z=0
\end{align}
where
\begin{align}
 k^2=k_{a\rho}^2+k_{az}^2\label{eqn:constrainteqncyltm}
\end{align}
which is the \emph{constraint equation} in cylindrical coordinates.

Since there are perfect electric conductors (pec) on the top and bottom of the cylinder, $Z_A(z)$ will be expanded as trigonometric functions which will form standing waves between the top and bottom plates.  The function and its derivative are defined as,
\begin{align}
Z_A(z) &= C_{ae}\cos(k_{az}z)+C_{ao}\sin(k_{az}z)\label{eqn:ZAcyl}\\
Z'_A(z) &= \frac{\partial{Z_A(z)}}{\partial{z}} = -k_{az}\left(C_{ae}\sin(k_{az}z)-C_{ao}\cos(k_{az}z)\right)\label{eqn:dZAcyl}
\end{align}

The fields in the $\hat{\phi}$ direction also form standing waves so they will be expanded as trigonometric functions in which it and its derivative are given as, 
\begin{align}
\Phi_A(\phi) &= D_{a0}\cos(\nu\phi)+D_{a1}\sin(\nu\phi)\nonumber\\
&=\sum_{\ell=0}^{1}D_{a\ell}\cos\left(\nu\phi-\ell\frac{\pi}{2}\right)\label{eqn:PhiAcyl}\\
\Phi'_A(\phi) &= \frac{d{\Phi_A(\phi)}}{d{\phi}} = -\nu\left(D_{a0}\sin(\nu\phi)-D_{a1}\cos(\nu\phi)\right)\nonumber\\
&= -\nu \sum_{\ell=0}^{1}D_{a\ell}\sin\left(\nu\phi-\ell\frac{\pi}{2}\right)\label{eqn:dPhiAcyl}
\end{align}
where in general $\nu$ is a real number.

Since the fields are assumed to be propagating in the $\hat{\rho}$ direction, $R_A(\rho)$ will be expanded as forward and backward traveling waves with constant coefficients, which are most conveniently described as \emph{Hankel} functions in the cylindrical coordinate system.  The function along with it's derivative are given as,
\begin{align}
R_A(\rho) &= R_A^+H_\nu^{(2)}(k_{a\rho}\rho)+R_A^-H_\nu^{(1)}(k_{a\rho}\rho)\label{eqn:RAcyl}\\
R_A'(\rho) &= \frac{d{R_A(\rho)}}{d{\rho}} = R_A^+{H_{\nu}^{(2)}}'(k_{a\rho}\rho)+R_A^-{H_{\nu}^{(1)}}'(k_{a\rho}\rho)\label{eqn:dRAcyl}
\end{align}
where 
\begin{align}
{H_{\nu}^{(2)}}'(k_{a\rho}\rho)&=\frac{d}{d\rho}[{H_{\nu}^{(2)}}(k_{a\rho}\rho)]=-k_{a\rho}H_{{\nu}+1}^{(2)}(k_{a\rho}\rho)+\frac{\nu}{\rho}H_{\nu}^{(2)}(k_{a\rho}\rho)\label{eqn:Hnu2A}\\
{H_{\nu}^{(1)}}'(k_{a\rho}\rho)&=\frac{d}{d\rho}[{H_{\nu}^{(1)}}(k_{a\rho}\rho)]=-k_{a\rho}H_{{\nu}+1}^{(1)}(k_{a\rho}\rho)+\frac{\nu}{\rho}H_{\nu}^{(1)}(k_{a\rho}\rho)\label{eqn:Hnu1A}
\end{align}
The variable $\nu$ is the order of the Hankel function and is the variable that couples $R_A(\rho)$ and $\Phi_A(\phi)$.

\subsection{$TM_z$ Field Solutions}
For $TM_z$ modes, the electric and magnetic fields are found by substituting (\ref{eqn:Azcyl}) into (\ref{eqn:EnormTM})--(\ref{eqn:HperpTM}) and expanding in the cylindrical coordinate system to get,
\begin{multline}\label{eqn:ETMcyl}
\vec{E}_A(\rho,\phi,z) =  C_a\sqrt{Z_a}\frac{1}{k_{a\rho}}\biggl[R'_A(\rho)\Phi_A(\phi)Z'_A(z)\hat{\rho}\\+\frac{1}{\rho}R_A(\rho)\Phi'_A(\phi)Z'_A(z)\hat{\phi}+k_{a\rho}^2R_A(\rho)\Phi_A(\phi)Z_A(z)\hat{z}\biggr]
\end{multline}
\begin{align}
\vec{H}_A(\rho,\phi,z) &= \frac{C_a}{\sqrt{Z_a}}\left[\frac{1}{\rho}R_A(\rho)\Phi'_A(\phi)Z_A(z)\hat{\rho}-R'_A(\rho)\Phi_A(\phi)Z_A(z)\hat{\phi}\right]\label{eqn:HTMcyl}
\end{align}
where $'=\frac{\partial}{\partial{u}}$ with $u=\{\rho,\phi,z\}$. Solving for $k_{a\rho}$ in (\ref{eqn:constrainteqncyltm}) yields,
\begin{align}
k_{a\rho{n}} = -j\sqrt{k_{az}^2-k^2}\label{eqn:karho}
\end{align}
in which the negative sign was chosen deliberately on the square root to ensure waves whose magnitude approach zero in the direction of propagation.
\subsection{Boundary Conditions}
The field variations in the $\phi$ direction are periodic such that,
\begin{align}
\Phi_A(\phi)=\Phi_A(\phi+2\pi)
\end{align}
Since the fields are periodic then the fields must repeat every $2\pi$ so that $\nu=m$ where,
\begin{align}
m= \left\{ \begin{array}{ll}
0,1,2,\ldots & \textrm{if $\ell=0$}\\
1,2,3,\ldots & \textrm{if $\ell=1$}\\
\end{array} \right.
\end{align}
which can also be defined with the intermediate variable $m'$ as
\begin{align}
m'&= 0,1,2,\ldots\\
m &= m'+\ell
\end{align}

The boundary conditions in the $\hat{z}$ direction will now be applied to solve for $k_{az}$ and either $C_{ae}$ or $C_{ao}$. The first boundary condition applied will be to the bottom plate of the cylinder where
\begin{align}
E^A_{\rho}(z=0)=E^A_{\phi}(z=0)=0
\end{align}
which from (\ref{eqn:ETMcyl}) implies that 
\begin{align}
Z'_A(0) &= -k_{az}\left(C_{ae}\sin(k_{az}0)-C_{ao}\cos(k_{az}0)\right)=-k_{az}C_{ao}=0
\end{align}
and since $k_{az} \ne 0$ then $C_{ao}=0$.
Application of the boundary condition on the top plate
\begin{align}
E^A_{\rho}(z=z_0)=E^A_{\phi}(z=z_0)=0
\end{align}
implies that 
\begin{align}
Z'_A(z_0) &= -k_{az} C_{ae}\sin(k_{az}z_0)=0
\end{align}
and since $k_{az}\ne0$ and $C_{ae} \ne 0$ then $k_{az}z_0={n}\pi$, or
\begin{align}
k_{az} = \frac{{n}\;\pi}{z_0}
\end{align}
where ${n}=0,1,2,\ldots\infty$.  Equation (\ref{eqn:Zcyl}) and (\ref{eqn:dZcyl}) now reduce to,
\begin{align}
Z_A(z)_n &= \cos\left(k_{azn} z\right)\label{eqn:ZTMcyl2}\\
Z'_A(z)_n &= -k_{azn}\sin\left(k_{azn} z\right)\label{eqn:dTMZcyl2}
\end{align}
where $C_{ae}$ was set to unity. Also the separated functions (\ref{eqn:PhiAcyl}) and (\ref{eqn:dPhiAcyl}) change to,
\begin{align}
\Phi_A(\phi)_{\ell,m} &=\cos\left(m\phi-\ell\frac{\pi}{2}\right)\label{eqn:PhiAcyl2}\\
\Phi'_A(\phi)_{\ell,m}&=-m\sin\left(m\phi-\ell\frac{\pi}{2}\right)\label{eqn:dPhiAcyl2}
\end{align}
in which $C_{fon}=D_{f\ell,m}=1$.
Equations (\ref{eqn:RAcyl}) and (\ref{eqn:dRAcyl}) become,
\begin{align}
R_A(\rho)_{\ell,m,n} &= R_{A\ell,m,n}^+H_{m}^{(2)}(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-H_{m}^{(1)}(k_{f\rho{n}}\rho)\label{eqn:RAcyl2}\\
R'_A(\rho)_{\ell,m,n} &= R_{A\ell,m,n}^+{H_{m}^{(2)}}'(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-{H_{m}^{(1)}}'(k_{a\rho{n}}\rho)\label{eqn:dRAcyl2}
\end{align}
where 
\begin{align}
{H_{m}^{(2)}}'(k_{a\rho n}\rho)&=\frac{d}{d\rho}[{H_{m}^{(2)}}(k_{f\rho{n}}\rho)]=-k_{a\rho n}H_{m+1}^{(2)}(k_{a\rho n}\rho)+\frac{m}{\rho}H_{m}^{(2)}(k_{a\rho n}\rho)\label{eqn:dHm2cyl2A}\\ 
{H_{m}^{(1)}}'(k_{a\rho n}\rho)&=\frac{d}{d\rho}[{H_{m}^{(1)}}(k_{f\rho{n}}\rho)]=-k_{a\rho n}H_{m+1}^{(1)}(k_{a\rho n}\rho)+\frac{m}{\rho}H_{m}^{(1)}(k_{a\rho n}\rho)\label{eqn:dHm1cyl2A}
\end{align}

\subsection{Basis Set and Field Expansion}
After applying boundary conditions, the solutions are seen to be a discrete set of modes, and because of linearity, the total electric and magnetic fields are found by summing (i.e., superposition) each of the individual modes to get
\begin{multline}\label{eqn:EAcyl}
\vec{E}_A(\rho,\phi,z)=\sum_{{\ell}=0}^1 \sum_{{m}=0}^\infty \sum_{{n}=0}^\infty
C_{a\ell,m,n}\sqrt{Z_{an}}\frac{1}{k_{a\rho{n}}}
\biggl[R'_A(\rho)_{\ell,m,n}\Phi_A(\phi)_{\ell,m}Z'_A(z)_{n}\hat{\rho}\\
+\frac{1}{\rho}R_A(\rho)_{\ell,m,n}\Phi'_A(\phi)_{\ell,m}Z'_A(z)_{n}\hat{\phi}+
k_{a\rho{n}}^2R_A(\rho)_{\ell,m,n}\Phi_A(\phi)_{\ell,m}Z_A(z)_{n}\hat{z}\biggr]
\end{multline}
\begin{multline}\label{eqn:HAcyl}
\vec{H}_A(\rho,\phi,z)=\sum_{{\ell}=0}^1\sum_{{m}=0}^\infty\sum_{{n}=0}^\infty\frac{C_{a\ell,m,n}}{\sqrt{Z_{an}}}\biggl[\frac{1}{\rho}R_A(\rho)_{\ell,m,n}\Phi'_A(\phi)_{\ell,m}Z_A(z)_n\hat{\rho}\\-R'_A(\rho)_{\ell,m,n}\Phi_A(\phi)_{\ell,m}Z_A(z)_n\hat{\phi}\biggr]
\end{multline}
Each function and variable was tagged with an index, where each individual mode will have a triple of indices $\ell,m,n$. Therefore, a mode is distinguished from all other modes by this index triple. From (\ref{eqn:EAcyl}) and (\ref{eqn:HAcyl}) a basis set is defined, where the $\hat{\rho}$ components are given by
\begin{align}
e^A_\rho(\phi,z)_{\ell,m,n} &= -C_{a\ell,m,n}\sqrt{Z_{an}}\Phi_A(\phi)_{\ell,m}Z'_A(z)_{n}\\ 
h^A_\rho(\rho,\phi,z)_{\ell,m,n}&=\frac{C_{a\ell,m,n}}{\rho\sqrt{Z_{an}}}\;\Phi'_A(\phi)_{\ell,m}Z_A(z)_n
\end{align}
and where the tangential components are defined as
\begin{multline}
\vec{e}_A(\rho,\phi,z)_{\ell,m,n} = C_{a\ell,m,n}\sqrt{Z_{an}}\biggl[\frac{1}{k_{a\rho{n}}\rho}\Phi'_A(\phi)_{\ell,m}Z'_A(z)_{n}\hat{\phi}\\+k_{a\rho{n}}\Phi_A(\phi)_{\ell,m}Z_A(z)_{n}\hat{z}\biggr]\label{eqn:EAbasiscyl}
\end{multline}
\begin{align}
\vec{h}_A(\phi,z)_{\ell,m,n} &= \frac{C_{a\ell,m,n}}{\sqrt{Z_{an}}}k_{a\rho n}\Phi_A(\phi)_{\ell,m}Z_A(z)_n\hat{\phi}\label{eqn:HAbasiscyl}
\end{align}
The total field components normal to the $\hat{\rho}$ direction, expanded in terms of the basis set are given as,
\begin{align}
E^A_\rho(\rho,\phi,z)&=\sum_{\ell,m,n}e^A_\rho(\phi,z)_{\ell,m,n}\left[R_{A\ell,m,n}^+{H_{m}^{(2)}}'(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-{H_{m}^{(1)}}'(k_{a\rho{n}}\rho)\right]\\
H^A_\rho(\rho,\phi,z)&=\sum_{\ell,m,n}h^A_\rho(\phi,z)_{\ell,m,n}\left[R_{A\ell,m,n}^+{H_{m}^{(2)}}(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-{H_{m}^{(1)}}(k_{a\rho{n}}\rho)\right]
\end{align}
where the tangential fields expanded in terms of the basis set are given as,
\begin{align}
\vec{E}^A_{\bot}(\rho,\phi,z) = \sum_{\ell,m,n}\vec{e}_A(\rho,\phi,z)_{\ell,m,n}\left[R_{A\ell,m,n}^+{H_{m}^{(2)}}(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-{H_{m}^{(1)}}(k_{a\rho{n}}\rho)\right]\label{eqn:EAperpcyl}\\
\vec{H}^A_{\bot}(\rho,\phi,z) = \sum_{\ell,m,n}\vec{h}_A(\phi,z)_{\ell,m,n}\left[R_{A\ell,m,n}^+{H_{m}^{(2)}}'(k_{a\rho{n}}\rho)+R_{A\ell,m,n}^-{H_{m}^{(1)}}'(k_{a\rho{n}}\rho)\right]\label{eqn:HAperpcyl}
\end{align}
It should be noted that since the tangential electric fields are dependent on the variable $\rho$, or in other words $\rho$ cannot be factored out, then these basis fields are unfortunately not suitable for a scattering matrix formulation.

\subsection{$TM_z$ Radial Symmetry (Vertical Polarization)}
If the structure has radial symmetry then
\begin{align}
\Phi'_A(\phi)=0
\end{align}
which makes 
\begin{align}
\Phi_A(\phi) = c
\end{align}
where $c$ is a constant which will be set to unity.  Under these conditions the index $m$ and $\ell$ reduce to $m=\ell=0$.  
The other two separated variables reduce to,
\begin{align}
Z_A(z)_n &= \cos\left(k_{azn}z\right)\\
Z'_A(z)_n &= -k_{azn}\sin\left(k_{azn}z\right)
\end{align} 
\begin{align}
R_A(\rho)_{n} &= R_{An}^+H_0^{(2)}(k_{a\rho{n}}\rho)+R_{An}^-H_0^{(1)}(k_{a\rho{n}}\rho)\\
R'_A(\rho)_{n} &= -k_{a\rho{n}}\left[R_{An}^+{H_1^{(2)}}(k_{a\rho{n}}\rho)+R_{An}^-{H_1^{(1)}}(k_{a\rho{n}}\rho)\right]
\end{align}
The basis vectors reduce to,
\begin{align}
e^A_{\rho}(z)_n&=-C_{an}\sqrt{Z_{an}}Z'_A(z)_{n}=C_{an}\frac{n\pi}{z0}\sqrt{Z_{an}}\sin\left(\frac{n\pi}{z0}z\right)\\
h^A_{\rho}(z)_n&=0
\end{align}
and where the transverse components simplify to,
\begin{align}
\vec{e}_A(z)_{n} &=C_{an}\sqrt{Z_{an}}k_{a\rho{n}}Z_A(z)_{n}\hat{z}=C_{an}\sqrt{Z_{an}}k_{a\rho{n}}\cos\left(\frac{n\pi}{z0}z\right)\hat{z}\label{eqn:EAbasiscylradialsym}\\
\vec{h}_A(z)_{n} &= \frac{C_{an}}{\sqrt{Z_{an}}}k_{a\rho n}Z_A(z)_n\hat{\phi}=\frac{C_{an}}{\sqrt{Z_{an}}}k_{a\rho n}\cos\left(\frac{n\pi}{z0}z\right)\hat{\phi}\label{eqn:HAbasiscylradialsym}
\end{align}
where
\begin{align}
C_{an} = \left[{z_0}k_{a\rho{n}}^2\pi\left(1+\delta_{n,0}\right)\right]^{-\frac{1}{2}}
\end{align}
It is noted that since the magnetic field has no $\rho$ component, it is $TM_\rho$.  Additionally if the index $n=0$ then $E_\rho=0$ and the mode is $TEM_\rho$ and the electric fields are vertically polarized or polarized in the $\hat{z}$ direction. 

The total field components normal to the $\hat{\rho}$ direction, expanded in terms of the basis set reduce to,
\begin{align}
E^A_\rho(\rho,z)&=\sum_{n}e^A_\rho(z)_{n}\left[R_{An}^+{H_{1}^{(2)}}(k_{a\rho{n}}\rho)+R_{An}^-{H_{1}^{(1)}}(k_{a\rho{n}}\rho)\right]\\
H^A_\rho(\rho,z)&=0
\end{align}
where the tangential fields expanded in terms of the basis set are given as,
\begin{align}
\vec{E}^A_{\bot}(\rho,z) = \sum_{n}\vec{e}_A(z)_{n}\left[R_{An}^+{H_{0}^{(2)}}(k_{a\rho{n}}\rho)+R_{An}^-{H_{0}^{(1)}}(k_{a\rho{n}}\rho)\right]\label{eqn:EAperpcyl_sym}\\
\vec{H}^A_{\bot}(\rho,z) = \sum_{n}\vec{h}_A(z)_{n}\left[R_{An}^+{H_{1}^{(2)}}(k_{a\rho{n}}\rho)+R_{An}^-{H_{1}^{(1)}}(k_{a\rho{n}}\rho)\right]\label{eqn:HAperpcyl_sym}
\end{align}
